proof of Schroeder-Bernstein theorem using Tarski-Knaster theorem

Proof.

Suppose f:S→T and g:T→S are injective. Define a functionφ:P⁢(S)→P⁢(S)
by φ⁢(X)=S∖g⁢(T∖f⁢(X)).

If X⊆Y⊆S, then S∖g⁢(T∖f⁢(X))⊆S∖g⁢(T∖f⁢(Y)), and so φ is monotone. Since P⁢(S) is a complete lattice, we may apply the Tarski-Knaster theorem to conclude that the set of fixed points of φ is a complete lattice and thus nonempty.

Let C be a fixed point of φ. We have

S∖C=g⁢(T∖f⁢(C)).

Hence g|T∖f⁢(C):T∖f⁢(C)→S∖C
and f|C:C→f⁢(C) are bijections. We can therefore construct the desired bijection h:S→T by defining

h⁢(x)={f⁢(x)if ⁢x∈C(g|T∖f⁢(C))-1⁢(x)if ⁢x∉C.∎

The usual proof of Schroeder-Bernstein theorem explicitly constructs a fixed point of φ.